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Let $U$ be the set of unipotent upper triangular matrices and $B$ the upper triangular matrices of $GL_3$. How could I describe $GL_3/U$ ? Using coordinates, in a projective or an affine space.

For example, I already know the identification of $SL_2/U$ with $\mathbb{A}^2 \setminus (0,0)$ and the identification of $SL_2/B$ with $\mathbb{P}^1$ ($B,U$ the standard Borel of $SL_2$ and its unipotent radical).

Thank you.

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  • $\begingroup$ GL/B - it is "flag variety". Do you really need its description in coordinates ? It is projective variety, however no one usually thinks about in terms of explicit equations... The open dense cell is lower triangular matrices - so you can easyly get coordiantes on this cell – $\endgroup$ Jan 31, 2012 at 17:38
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    $\begingroup$ SL_2/U is not A^2, it is A^2 minus the origin. $\endgroup$ Jan 31, 2012 at 17:59
  • $\begingroup$ You're both right, I knew it is a projective variety, but as I'm trying to work out a proof in the setting of G= GL_3, I'm willing to explicit the construction with coordinates. $\endgroup$
    – th.ng
    Jan 31, 2012 at 18:34
  • $\begingroup$ Notice that $B = \mathbb{C}^\times \cdot U$ and so $G/B = (G/U)/\mathbb{C}^\times$. This is one way to see that $SL_2/U = \mathbb{A}^2 - (0,0)$ given that $SL_2/B$ is $\mathbb{P}^1$. I imagine you should also be able to realize $GL_3/U$ as the affine cone over the flag variety $GL_3/B = SL_3/(B\cap SL_3)$. One way to realize the latter as a projective variety is to look at the adjoint action of $SL_3$ on its Lie algebra and look at the orbit of a highest weight. $\endgroup$
    – solbap
    Jan 31, 2012 at 19:04

3 Answers 3

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Let $V$ be the basic (3-dimensional) representation of $GL(3)$. Then $SL(3)/U$ is the set of all pairs $x\in V, y\in V^*$ where $x$ and $y$ are non-zero and $(x,y)=0$.

The quotient $GL(3)/U$ is non-canonically product of the above by $C^{\times}$. Canonically, you need to choose non-zero $x_i\in \Lambda^i(V)$ (for $i=1,2,3$) such that $x_i\wedge x_j=0$ for all $i$ and $j$ (note that if $x_3$ is fixed then $\Lambda^2(V)$ is canonically the same as $V^*$). This description generalizes immediately to any $GL(n)$ (and in fact to any $G$).

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  • $\begingroup$ I don't understand the first identification of $SL_3/U$, could explain it ? Moreover, do you have some kind of references for the generalization to $GL_n$ or any $G$ ? $\endgroup$
    – th.ng
    Jan 31, 2012 at 20:23
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    $\begingroup$ Well, you just consider a pair (vector,covector) whose pairing is 0 (both should be assumed to be non-zero, otherwise you get not $SL(3)/U$ but its affine closure). There so many references that I am not even sure where to refer you... The statement is that to specify a point of $G/U$ (for simply connected $G$) you must specify a vector in every fundamental representation of $G$ subject to certain quadratic relations (called Plucker relations). For $GL(n)$ the fundamental representations are the wedge powers of the basic $n$-dimensional representation. $\endgroup$ Jan 31, 2012 at 20:45
  • $\begingroup$ The identification $SL_3/U$ with {(vector $\vec v$, 2-tensor $\tau$) : $\vec v \wedge \tau = 0$} is given by $M \mapsto$ (first column of $M$, first two columns wedged together). $\endgroup$ Feb 1, 2012 at 1:35
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As already mentioned, the quotient $GL_n(\mathbb{C})/B$ is the space $Fl$ of all flags ($0\subset$ a line $\subset$ a 2-plane $\subset\cdots\subset \mathbb{C}^n$). There are line bundles $L_i , i=1,\ldots, n$ on this space where $L_i$ is obtained by pulling back the tautological bundles from Grassmannians of $i$- and $i-1$-planes in $\mathbb{C}^n$ and quotienting one by the other.

Let $L_i^0$ be the total space of $L_i$ minus the zero section and set $L^0$ to be the fibered product $$L_1^0\times_{Fl}L_2^0\times_{Fl}\cdots\times_{Fl}L^0_n.$$ The group $GL_n(\mathbb{C})$ acts transitively on the space $L^0$ and each stabilizer is conjugate to the group of unipotent matrices in $GL_n(\mathbb{C})$. So $GL_n(\mathbb{C})/\mbox{unipotent matrices}\cong L_0$.

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Let $G$ over a field $k$ be a reductive group and $P$ be parabolic subgroup with Levi decomposition $P =MU$ (over $k$), then $$G / U = \amalg_{w \in N(M) /M} UMwU,$$ where $N_G(M)$ is the normalizer of $M$ in $G$.

Since $MU = UM$ and $Mw = wM$ and $wM \cap M = wU \cap U = \emptyset$ for $w \neq 1$, we encounter that $$ G/U = M \amalg \; \coprod\limits_{w\neq 1} UMw$$ and $$G/P = 1 \amalg \;\coprod\limits_{w\neq 1} Uw.$$

I hope this is clearer now.

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  • $\begingroup$ Actually, I was looking for a more explicit identification, for example, I was able to embed $GL_3/B$ in $\mathbb{P}^2 \times \mathbb{P}^2$, so I was looking for something like that. But I'll see if I can get something from the Bruhat decomposition. Thank you. $\endgroup$
    – th.ng
    Jan 31, 2012 at 18:37

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